1. Tardigrade
  2. Question
  3. Physics
  4. A particle located in a one-dimensional potential field has its potential energy function as U(x) (a/x4) - (b/x2)where a and b are positive constants. The position of equilibrium x corresponds to

Question Error Report

Thank you for reporting, we will resolve it shortly

Back to Question

Q. A particle located in a one-dimensional potential field has its potential energy function as $U\left(x\right) \frac{a}{x^{4}} - \frac{b}{x^{2}}$where $a$ and $b$ are positive constants. The position of equilibrium $x$ corresponds to

Work, Energy and Power

Solution:

The position of equilibrium corresponds to $F\left(x\right) = 0$
Since $F\left(x\right) = \frac{-dU\left(x\right)}{dx}$
so $F\left(x\right) = -\frac{d}{dx}\left(\frac{a}{x^{4}} -\frac{b}{x^{2}}\right)$ or$ F\left(x\right) = \frac{4a}{x^{5}} -\frac{2b}{x^{3}}$
For equilibrium, $F\left(x\right) = 0$, therefore
$\frac{4a}{x^{5}} - \frac{2b}{x^{3}} = 0$
$\Rightarrow x=\pm \sqrt{\frac{2a}{b}}$
$\frac{d^{2}U\left(x\right)}{dx^{2}} = - \frac{20a}{x^{6}} + \frac{8b}{x^{4}}$
Putting $x = \pm \sqrt{\frac{2a}{b}}$ given $\frac{d^{2}U\left(x\right)}{dx^{2}}$ as negative
So $U$ is maximum. Hence, it is position of unstable equilibrium